Advanced Placement (AP), 22.07.2019 08:20, organicmemez

Short form of writing is becoming a standard in the medical workplace. explain why brevity in writing is useful in the in the healthcare industry.

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Brevity is useful bevause it simply saves time - if the writing was general, healthcare workers would need to consult the writer as to what they mean specifically. Health professionals need to know the exact illness or side effects in order to properly care for their patients

answered: wonderland12372

Brevity in writing is useful in the in the healthcare industry because it is a time saverFurther explanation

Short form of writing is becoming a standard in the medical workplace, because it is easier for the busy healthcare workers, so they don't have to write as much and it is much quicker to write abbreviations

Brevity is defined as the shortness of time or duration I think brevity is important in a healthcare industry because it is a time saver. Each step in healthcare industry spent by getting information from the patient. By using short solution it or brevity will get information down so the rest of the time can be spent on taking care of other important things. Brevity can saves time, and time saves lives. Because especially in healthcare industry we dont want to have to say a twenty four letter words when you can use three to save time and save lives of the patients.

Learn more Learn more about brevity definition linkLearn more about healthcare industry linkLearn more about working in healthcare industry linkLearn more about the Short form of writing linkAnswer details

Grade: 9

Subject: Advanced Placement (AP)

Chapter: the medical workplace.

Keywords: the medical workplace.

answered: Guest

The answer is (d) 7/2======we have $f(x) = \begin{cases} x & \text{ for } x \le 0 \\ x + 1 & \text{ for } x \ \textgreater \ 0 \end{cases}$ we also have an integral $\displaystyle\int_{-2}^1 x\, f(x)\; dx$ notice how function f splits up at x = 0. there's a left piece and a right pieceso we'll need to do the same thing with our integral: split it at x = 0we use a property of integrals that allows us to split up and sum: $\displaystyle\int_{-2}^1 x\, f(x)\; dx = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx$ notice that for the piecewise function f, the function has the equationf(x) = x when x ≤ 0. so for the integral $\int_{-2}^0 x\, f(x)\; dx$ , f(x) = x so it changes into $\begin{aligned} \int_{-2}^0 x\, f(x)\; dx & = \int_{-2}^0 x \cdot x\; dx \\ & = \int_{-2}^0 x^2\; dx \\ & = \left[\frac{x^3}{3}\right]_{-2}^0 & & \text{\footnotesize antidifferentiate} \\ & = 0 - \frac{(-2)^3}{3} \\ & = \frac{8}{3} \end{aligned}$ for the piecewise function f, the function has the equationf(x) = x+1 when x > 0. for the integral $\int_{0}^1 x\, f(x)\; dx$ , we have $\begin{aligned} \int_{0}^1 x\, f(x)\; dx & = \int_{0}^1 x(x+1)\; dx \\ & = \int_{0}^1 (x^2+x)\; dx \\ & = \left[ \frac{x^3}{3} + \frac{x^2}{2}\right]_0^1 \\ & = \frac{1}{3} + \frac{1}{2} - 0 = \frac{5}{6} \end{aligned}$ therefore, back to our splitted integral: $\begin{aligned} \displaystyle\int_{-2}^1 x\, f(x)\; dx & = \displaystyle\int_{-2}^0 x\, f(x)\; dx + \displaystyle\int_{0}^1 x\, f(x)\; dx \\ & = \frac{8}{3} + \frac{5}{6} \\ & = \frac{7}{2} \end{aligned}$
Can anyone me on this ap calc question? ?